Conjecture: Every discontinuous utility function $U$ representing continuous preferences can be written as $U = f \circ g$ for some continuous $g$ and discontinuous strictly monotone $f$.

The goal is to prove or disprove this conjecture.

We know that continuous utility implies continuous preferences, and running it through a strict monotone gives us the same preferences but with a discontinuous utility function. The conjecture is inspired from this.

  • $\begingroup$ I recommend that you delete your counter-example and the edit from the body of the Q. $\endgroup$
    – Giskard
    Apr 7, 2022 at 14:12
  • $\begingroup$ Are the continuous preferences themselves monotone? $\endgroup$
    – Giskard
    Apr 7, 2022 at 14:14
  • $\begingroup$ @Giskard The preferences need not be monotone. (I have edited the post as well.) $\endgroup$ Apr 7, 2022 at 14:16
  • $\begingroup$ Any assumptions on the domain of $U$? $\endgroup$ Apr 7, 2022 at 14:36
  • $\begingroup$ @MichaelGreinecker The domain $X \subseteq \mathbb{R}^n$. But I would like to see these two cases, if possible: (1) $X \subseteq \mathbb{R}^n$ and (2) $X = \mathbb{R}^n$. $\endgroup$ Apr 7, 2022 at 14:57

2 Answers 2


I took the question as asking for a representation with $f:\mathbb{R}\to\mathbb{R}$. If we only require $f$ to be defined on a subset of $\mathbb{R}$, Amit's answer solves the problem.

Here is a proof for the case that the domain is connected (and second countable): Under these assumptions, there exists a continuous utility representation $V:X\to\mathbb{R}$. Since continuous functions map connected sets to connected sets, the image $V(X)$ is a, possibly unbounded, interval. I do the special case that $V(X)=[a,b)$ for some real numbers $a$ and $b$, the other cases can be dealt with by analogous methods. Now define the surjective continuous function $g:X\to [a,\infty)$ by $$g(x)=\frac{V(x)-a}{b-V(x)}.$$ Note that $g$ represents the preference ordering. Let $r^*=U(x)$ for any (and hence all) $x\in g^{-1}(a).$

Now define $f:\mathbb{R}\to\mathbb{R}$ so that for $r\geq a$ we have $f(r)=U(x)$ for any (and hence all) $x\in g^{-1}(r)$ and for $r<a$ we have $f(r)=r^*-|r-a|$. It is easy to verify that $f$ is strictly increasing and $U=f\circ g$.

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – 1muflon1
    Apr 10, 2022 at 12:36

If the preference is continuous, reflexive, transitive and complete, then there exists a continuous utility representation $g$, and since $U$ also represents the same preference, so there must be a strictly increasing function $f$ such that $U= f\circ g$.

  • $\begingroup$ Can you please elaborate the second part where you say that such an $f$ must exist? It may be true, but I can't convince myself that it really is the case. A proof would be appreciated! $\endgroup$ Apr 7, 2022 at 16:30
  • $\begingroup$ Take it as an exercise. Try to construct such a $f$, it is easy. $\endgroup$
    – Amit
    Apr 7, 2022 at 17:03
  • $\begingroup$ The existence of $g$ comes from Debreu's theorem. Then define $f$ such that $f(g(x)) = U(x)$. Since $g$ and $U$ denote the same preferences, it follows that $f(g(x)) > f(g(y)) \iff U(x) \geq U(y) \iff g(x) \iff g(y)$ which further implies that $f$ is monotone. Does this work? $\endgroup$ Apr 8, 2022 at 0:18
  • $\begingroup$ $f$ defined as $f(g(x)) = U(x)$ works provided you want $f:\text{Range}(g)\rightarrow\mathbb{R}$. Here $\text{Range}(g)$ will be an interval because $g$ is continuous. If you want $f:\mathbb{R}\rightarrow\mathbb{R}$, then you just need to complete the construction of $f$ outside the range of $g$. $\endgroup$
    – Amit
    Apr 8, 2022 at 2:09
  • $\begingroup$ @Pocket Cat If you want $f:\mathbb{R}\rightarrow\mathbb{R}$, see Michael's answer. $\endgroup$
    – Amit
    Apr 8, 2022 at 7:30

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